# An exercise in infinite combinatorics from Burris and Sankappanavar

Exercise 6.7 in chapter IV of Burris and Sankappanavar's A Course in Universal Algebra starts as follows:

Show that for $I$ countably infinite there is a subset $S$ of the set of functions from $I$ to $2$ which has cardinality equal to that of the continuum such that for $f \not= g$ with $f,g \in S$, $\{i \in I : f(i) = g(i)\}$ is finite.

I don't see how $S$ can have cardinality more than 2: let $f, g, h \in S$, with $f \not= g$ and $f \not= h$ and let $A = \{i \in I : f(i) = g(i)\}$ and $B = \{i \in I : f(i) = h(i)\}$. For $i$ in $I\backslash (A \cup B)$ we have $g(i) \not= f(i) \not= h(i)$, whence $g(i) = h(i)$, since $f(i), g(i), h(i) \in 2 = \{0, 1\}$. But $A$ and $B$ are finite, $I \backslash(A\cup B)$ is infinite. So $g(i) = h(i)$ holds for infinitely many $i$ and we must have $g = h$.

What have I missed? (A redefinition of 2, perhaps?)

I would also be grateful for a reference for what appears to be the goal of this exercise, namely to use an ultraproduct construction to obtain uncountable models from countable ones.

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Since the main goal is to show that $|\,A^I / U\,| \geq 2^{\omega}$ whenever $U$ is a nonprincipal ultrafilter on a countably infinite set $I$ and $A$ is an infinite set, it would make sense to instead find a family $S$ of continuum-many functions $I \to A$ such that $\{ i \in I : f(i) = g(i) \}$ is finite for all distinct $f,g \in S$. –  Arthur Fischer Aug 2 '12 at 11:12
Very likely they only want the sets $\{i \in I\,:\,f(i) = g(i) =1\}$ to be finite (at least from what @Arthur said). Such a family of functions is called almost disjoint. A number of nice arguments of the existence of such families is given in the second section of Geschke's notes on Almost disjoint and independent families. –  t.b. Aug 2 '12 at 11:28
I am sure Arthur is right that the first part of the exercise should be about functions $I \rightarrow A$ not $I \rightarrow 2$ (since the second part then follows easily). With this correction, the claim follows easily from the first lemma in Geschke's notes. My thanks to Arthur and t.b. both. –  Rob Arthan Aug 2 '12 at 12:05
Prof. Burris has corrected the exercise in the 2012 update to the book. –  Rob Arthan Feb 28 '13 at 23:04

On the website of the book there is a newer version and - as the OP already pointed out in their comment the exercise is already corrected there (see page 150). It says:

Show that, for $I$ countably infinite and $A$ infinite, there is a subset $S$ of the set of functions from $I$ to $A$ which has cardinality equal to that of the continuum such that for $f\ne g$ with $f,g\in S$, $\{i\in I; f(i)=g(i)\}$ is finite. Conclude that $|A^I/U|\ge 2^\omega$ if $U$ is a nonprincipal ultrafilter over $I$.


So we can take $I=\N$ and $A$ can be an arbitrary infinite countable set.

For $A=\Q$ we can repeat basically the same trick as the trick which is used to show the existence of an almost disjoint family of cardinality $\mathfrak c$. See, for example, here or here.

For any real number $r\in\R$ we take any injective sequence $f_r \colon \N \to \Q$ of rational numbers, which converges to $r$. The system $\{f_r; r\in\R\}$ has the same cardinality as $r$. And for any $r\ne r'$ the set $\{i\in\N; f_r(i)=f_{r'}(i)\}$ is finite, since these two sequences have different limits.

A different approach, which only shows that there is uncountably many such functions.

Lemma: If $\{f_n; n\in\N\}$ is a sequence of functions from $\N$ to $\N$ then there is a function $f$ such that each set $A_n=\{i\in\N; f(i)=f_n(i)\}$ is finite.

Proof. We can define $$f(n)=\max\{f_i(j); i=0,\dots,n-1; j=0,\dots,n-1\}+1.$$ Then we have $\{i\in\N; f_n(i)=f(i)\}\subseteq\{0,1,\dots,n\}$. $\qquad\square$

Using the above fact and transfinite induction we can get a family of functions with cardinality $\aleph_1$, which has the required properties.

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